Diode & Electronic Circuits

1. Diode & Electronic Circuits

2. Semiconductors • conductor • easily conducts electrical current • valence electron can easily become free electrons • insulator • does not conduct electrical current • valence electron are tightly bound to the atoms. • semiconductor • atoms are arranged in a crystal.

3. Semiconductors (cont.) • each atom shares its electrons with four neighboring atoms • the valence orbit can hold no more than eight electrons,

4. Electron and Hole Current • heat energy causes the atoms to vibrate • dislodge an electron from the valence band to the higher conduction band • departure of the electron creates a vacancy in the valence band called a hole • hole will attract and capture any electron in the immediate vicinity for recombination

5. Electronic Current

6. Hole Current

7. Doping a Semiconductor An intrinsic (pure) semiconductor has very few free electrons and holes. To increase the number of carrier (holes and electrons), doping is used, n-Type Semiconductor doped with a pentavalent impurity (donor atom) free electrons are called the majority carries p-Type Semiconductor doped with a trivalent impurity (acceptor atom) the holes are called the majority carriers

8. Free electron Hole (n type) (p type) Pentavalent dopant Trivalent dopant

9. PN Junction Diode half of a silicon crystal doped with p-type impurity ,another half doped with n type impurity the border between p-type and n type is called the pn junction p n Anode R = VS Cathode

10. Forward Bias anode is more positive than cathode (>0.7v, the knee voltage) diode is conduct with large current ID acts as a close switch neglect 0.7V in ideal diode Reverse Bias cathode is more positive than anode diode passes a negligible amount of reverse current acts as an open switch diode will breakdown when reverse voltage is too large

11. 200 175 150 125 100 Forward current in mA 75 50 25 knee 0 0 0.5 1.0 1.5 Forward bias in volts

12. breakdown Reverse bias in Volts 400 600 200 0 20 40 Reverse current in mA 60 80 100 120 140

13. Ideal diode act as switch

14. Rectification and Smoothing • The above diagram show how to obtain a regulated d.c. from the high voltage a.c. main • This section will discussed the rectification and smoothing

15. Vin Vout The half-wave rectifier Ideal: VP(in) = VP(out) Vin Vout

16. Halfwave Rectifier (cont.) • During the positive cycle of the a.c. source, • the diode act as a close switch (neglect 0.7V) • Vp(out) = Vp(in) = Vsec • During the negative cycle of the a.c. source • the diode act as an open switch • Output Frequency : fout = fin • D.C. value of Half-Wave Signal

17. Example 5.1 What are the peak load voltage and d.c. load voltage of fig. 5.8 ? Fig 5.8 Example of half wave rectifier

18. The peak load voltage = peak secondary voltage = 34V The dc load voltage = 0.318 V2p = 10.8V Solution Given the turns ratio is 5 : 1. The r.m.s. secondary voltage is one-fifth of the primary voltage • The peak secondary voltage is

19. Vin Vout The bridge rectifier D1 D2 D3 Vout D4

20. Full wave (Bridge) Rectifier • During the positive cycle, • the purple diode D2 and D3conduct, • the red diode D1 and D4do not conduct • During the negative cycle • the purple diode D2 and D3 do not conduct, • the red diode D1 and D4 conduct • Output Frequency • Half wave : fout =2 fin • D.C. value of Bridge Rectifier

21. Smoothing circuit • During the positive first quarter-cycle of the input, the diode is forward-biased, allowing the capacitor to charge to point a. • When the input begins to decrease below its peak, the diode becomes reverse-biased, the capacitor is discharged from point a to point b

22. Point b is the point in which Vin is larger than the Vc again. The capacitor resumes the charging again until it reaches point c (= point a) • Repeating the discharging from point a to point b and the charging from point b to point c, we obtain the following waveform.(it is much smooth than the waveform without capacitor.)

23. Transistor Transistor characteristics • A bipolar transistor is an electronic device made of materials such as silicon and germanium. • It consists of three layers, either P-N-P or N-P-N. • Transistor can be used as amplifier or switch. • In this section, only n-p-n transistor will be discussed.

24. The three terminals of the transistor are collector, base and emitter. • Under normal operating condition, the junction of collector and base should be reverse biased and the junction of base and emitter junction should be forward biased.

25. In the following diagram, the base of the transistor is connected to the common point, i.e., the junction of the two power supplies, this method is known as common base. • The methods of common emitter and common collector are also used in circuits. • In the common emitter configuration, the input is applied to the base and the output is taken from the collector.

26. Symbol of NPN transistor Arrangement of an NPN transistor

27. Transistor switch and amplifier • Under steady state, IB is kept constant. The current gain of the transistor is dc = IC / IB. • In the linear region, dc will not vary too much with IB. • The current flowing into the base and into the collector will flow out of the transistor from the emitter, therefore, IE = IC + IB or IE = (1+ dc)IB.

28. When the transistor is operating as an common emitter amplifier, the input is applied to the base and output is taken from the collector. • It should be biased to have VCE approximately equal to VCC/2 at the steady state. Then the amplifier will have widest dynamic range. • When an input signal applied, IB varies and IC will vary accordingly. • If the input is too large, IB will swing out of the linear region and the output will be distorted.

29. The relationship between the input and output voltages can be shown in the following formula: • Where Vin and Vout are input and output voltage respectively, G is the voltage gain of the amplifier.

30. Common emitter transistor amplifier

31. Amplifiers are classified in four types, they are classes A, AB, B and C. • They are distinguished by the conduction angle of collector current. • For class A amplifier, the whole input waveform will appear at the output. Its conduction angle is 360 . The transistor is only operating in linear region. • For class B amplifier, only half of the input waveform will appear at the output. Its conduction angle is 180.

32. The conduction angle of class AB amplifier is between that of class A and class B, less than 360 but larger than 180. • For class C, less than half of the waveform will appear at the output. Its conduction angle is less than 180.

33. Classes of amplifier

34. Among the four types, only class A amplifier does not have high distortion output. However, it has a draw back of low efficiency. Therefore, it is usually used on small signal amplifier. • Class AB can also work on analogue signals but at least two transistors will be used to build an amplifier to achieve low output distortion. It has higher efficiency than class A amplifier. • Because of high output distortion, class B and C amplifiers are seldomly seen operating on analogue signals. But their high efficiency make them favorable on radio frequency high power applications.

35. When the transistor is operating in the cutoff state, IB is very small or equal to zero. IC is very small too and VCE VCC. • When the transistor is operating in the saturation state, IB and IC are both high, then VCE 0. • A transistor amplifier will not have undistorted output when it is operating in cutoff or saturation state. • However, transistor behaves as a switch in this two states.

36. In the cutoff state, the transistor is said to be turned off, and the output voltage is high. • In the saturation state, the transistor is said to be turned on, and the output voltage is low. • Since the logic ICs can only deliver very small output current, the transistor switch can be used to boost the output current to drive the high power devices.

37. When the input is at high voltage level, its output is at low level and vice versa. This switch is also an inverter. • The applications include switch mode power supply, controlling electromechanical devices (eg. solennoid or motor) and high power electrical devices (eg. light bulb or heater).

38. Transistor switch

39. Example 1 An input of 10mV is applied to an amplifier with voltage gain 200. What is the output voltage? Output voltage Vout= GVin = 200 x 10mV = 2V

40. Example 2 In the circuit diagram shown in fig. 5.17, VCC =15V, RC=1k, dc =100. What is the most appropriate VCE? Calculate IC, IB, IE and RB under steady state. If the same transistor is used as a switch as on fig. 5.18 and the load resistor equals to 200, calculate RB.

41. Example 2 VCE = VCC / 2 = 7.5V IC = (VCC - VCE )/ RC = (15V-7.5V)/1000 = 7.5mA IB = IC/dc = 7.5mA/100 = 0.075mA or 75A IE = IC + IB = 7.5mA + 0.075mA =7.575mA Since VBE =0.7V, therefore, RB = (VCC - VBE )/ IB =(15V-0.7V)/0.075mA=191k

42. Example 2 When the transistor is used as a switch, RB of lower value should be chosen. When the switch is turned on, IC = VCC / RC = 15V/200 = 75mA IB should be higher than IC / dc = 75mA/100=0.75mA RB should be lower than (VCC - VBE )/ IB =(15V-0.7V)/0.75mA=19k To make sure that the transistor can turn on, RB should be chosen to be 1/10 of the above calculated value, that is 1.9k

43. A INPUT OUTPUT A 0 1 1 0 Logic Function NOT Function • It has one input and one output • The output Y is equal to the complement of A(input). • It is denoted by a prime (‘)after the alphabet A. • The logical operator is ( )and the operand is A.

44. It is important to point out that each input can only take one of the two logic values, 1 or 0 • To work out the output for each combinations of input, we have to use Boolean Algebra • Logic Gate is the equivalent electronic device that performs a Boolean Algebraic function.

45. A Y=A B ˙ B IN PUTS OUTPUT A B Y = A B ˙ 0 0 0 0 1 0 1 0 0 1 1 1 AND Function • Y is HIGH only when both A and B are HIGH. • Y = A•B (sometimes the dot • is omitted) stands for a Boolean operation, AND multipication. • AND multiplication is exactly the same as ordinary multiplication.

46. IN PUTS OUTPUT A B Y = A+B 0 0 0 0 1 1 1 0 1 1 1 1 OR Gate • Y is HIGH unless both input A, B are LOW. • Y = A+B stands for a Boolean operation, OR addition. • OR addition is similar to ordinary addition except that 1+1 = 1

47. Multiple Level of Gates B A C D • The output (1st level) gate is a 3 input AND gate • 2 inputs are input A,D • remaining input is fed from the output of an OR gate • the OR gate is called 2rd level gate • output Y is a product of three terms

48. A B C • The output gate is a 2 input OR gate • each input from output of AND gate • The two AND gates are called 2nd level gate • Output X is a sum of two terms

49. A B C 0 0 0 1 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 0 0 1 0 1 1 1 1 0 1 0 1 Truth Table for

50. A B C D B+C 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 0 0 1 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 Truth Table for